/sci/ - Science & Math

p14 ex17

We have that [math][L_1L_2:F]=[L_1L_2:L_1][L_1:F][/math], so it is enough to show that [math][L_1L_2:L_1]\leq [L_2:F][/math]. But this is obviously true since the minimal polynomial over [math]F[/math] of any element in [math]L_2[/math] is also a polynomial over [math]L_1[/math] of same degree.

The last part follows because each of [math][L_i:F][/math] are factors of [math][L_1L_2:F][/math], so together with the inequality shows the result.